Explicit description of SU(2,2)/U

Question

Consider the real diagonal $4\times 4$ - matrix
$$I_{2,2}={\rm diag}(1,1,-1,-1)$$ and the corresponding special unitary group $$ G={\rm SU}(2,2)=\{g\in {\rm SL}(4,{\mathbb{C}})\ |\ g\cdot I_{2,2}\cdot \bar g ^{\rm tr}=I_{2,2}\}.$$ We regard $G$ as an algebraic group over ${\mathbb{R}}$. It is known that $G$ is quasi-split, that is, $G$ contains a Borel subgroup $B$ defined over ${\mathbb{R}}$. Let $U$ denote the unipotent radical of $B$, it is defined over ${\mathbb{R}}$. Set $X=G/U$.

Question. How can one describe the homogeneous space $X$ explicitly (by equations and "inequalities") as a quasi-projective variety over ${\mathbb{R}}$ on which ${\rm SU}(2,2)$ naturally acts?

EDIT: I explain why I need explicit equations for $G/U$. I want to twist the desired variety by the 1-cocycle $c=I_{2,2}\in G(\mathbb{R})_2$ and to obtain explicit equations for the twisted variety $Y={}_c X$. This new variety $Y$ over $\mathbb{R}$ is a homogeneous space of the twisted group $_cG={\rm SU}(4)$ for which $U$ is the stabilizer of a $\mathbb C$-point. Of course, $Y$ has no $\mathbb R$-points, because the stabilizer of an $\mathbb R$-point would be a compact form of $U$, but we know that the unipotent group $U$ has no compact forms.

Answer 1

I guess your variety is just the variety of pairs of ${\mathbb C}$-linearly independent vectors in ${\mathbb C}^4$ that are isotropic with respect to this Hermitian form and orthogonal to each other. Respectively, the twisted form is the same variety but for another Hermitian form (and if the form is not hyperbolic, there is no such a pair over ${\mathbb R}$, as you mentioned).

Answer 2

Although the question has been answered in comments (by Victor Petrov), I prefer to post an answer. I assume that $G={\rm U}(2,2)$ rather than $G={\rm SU}(2,2)$.

1. My variety $G/U$ is the variety $X$ whose real points are the triples $$(W,w,b),$$ where $W\subset \mathbb C^4$ is an isotropic 2-dimensional subspace, $w\in W$ a nonzero vector (which is automatically isotropic), and $b$ is a nonzero element of $\Lambda^2W$. This variety is a $R_{\mathbb C/\mathbb R}\mathbb G_{m,\mathbb C}^2$-torsor over the variety $\mathcal F$ of isotropic flags: the map is $$X\to\mathcal F\colon\quad (W,w,b)\mapsto (W,\langle w\rangle),$$ and the action of $(\mathbb C^\times)^2$ on $X$ is $$ (\lambda,\mu)*(W,w,b)=(W,\lambda w,\lambda\mu b)\quad \text{for } \lambda,\mu\in\mathbb C^\times.$$ By Witt's theorem for Hermitian forms, $G(\mathbb R)$ transitively acts on $X(\mathbb R)$, and my calculations show that the stabilizer of the point $$(\langle e_1,e_2\rangle, e_1, e_1\wedge e_2)\in X(\mathbb R)$$ is a maximal unipotent subgroup of $G$. Thus $X\simeq G/U$. The twisted form of $X$ is the same variety, but for another Hermitian form (and if the form is not hyperbolic, there is no such triples $(W,w,b)$ over $\mathbb R$, as Victor has mentioned).

2. The real points of the variety $\mathcal V$ of Victor's answer are pairs of non-proportional isotropic vectors $(w_1,w_2)$ in $\mathbb C^4$. This variety is a $R_{\mathbb C/\mathbb R}\mathbb G_{a,\mathbb C}$-torsor over $X$: the map is $$\mathcal V\to X\colon\quad (w_1,w_2)\mapsto (\,\langle w_1,w_2\rangle,\, w_1,\, w_1\wedge w_2)$$ and the action of $\mathbb C$ on $\mathcal V$ is $$a*(w_1,w_2)=(w_1, w_2+aw_1)\quad\text{for } a\in\mathbb C.$$

3. The real points of the variety $\mathcal V'$ of Victor's comment is the set of pairs $(v,u)$, where $v$ is a nonzero isotropic vector in $\mathbb C^4$, and $u$ is a nonzero vector in $v^\perp/\langle v \rangle$ that is isotropic with respect to the induced Hermitian form on $v^\perp$. We construct a $G$-equivariant isomorphism $$\varphi\colon\mathcal V'\to X.$$ Let $(v,u)\in \mathcal V'(\mathbb R)$. We lift $u$ to an isotropic vector $\tilde u\in \mathbb C^4$ and set $$\varphi(v,u)=(\langle v, \tilde u\rangle, v, v\wedge\tilde u)\in X(\mathbb R).$$ In the opposite direction, if we have $(W,w,b)\in X(\mathbb R)$, we choose $y\in W$ such that $b=w\wedge y$, and we set $$\psi(W,w,b)=(w, y+\langle w\rangle)\in\mathcal V'(\mathbb R).$$ Since $\varphi$ and $\psi$ are mutually inverse, we see that $\varphi$ is an isomorphism.